CE 573: Sediment Transport

HOME WORK 1

Professor:Dr. Weiming Wu

By Cássio Rampinelli

January, 31th, 2019

OBS: This script was done in R programming language

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PROBLEM 2.3. Why quartz is the most abundant mineral in riverine sediments?

Solution

Quartz is a mineral composed of silicon and oxygen atoms. These atoms are one of the most common elements in the Earth crust which in part explains the abundance of Quartz in riverine sediments. In addition, the atoms structural arrangement provides a physical/chemical framework that makes quartz relatively resistant to chemical weathering.

 

PROBLEM 2.8. The sediment concentration by volume is measured as 0.1. Assume the sediment specific gravity is 2.65.

 

1) Determine the sediment concentration by weight, by mass per unit volume, and by ppm by weight.

Solution

 

Given:

\(c=0.1\) is the sediment concentration by volume;

\(G=2.65\) is the specific gravity;

\(\rho_{f}= 1,000 kg/m^3\) is the water density at 4 celsius degrees;

 

Sediment concentration by weight (\(\hat{c}\)):

The correlation between the concentration by volume (\(c\)), and the concentration by weight (\(\hat{c}\)) is given by:

\[ \hat{c}=\frac{Gc}{1+(G-1)c} \] \[ \hat{c}=\frac{2.65 \cdot 0.1 }{1+(2.65-1)0.1} \] \[ \hat{c}=0.227 \]  

Sediment concentration by mass per unit volume (\(c_{mass}\))

Considering \(\rho_{s}\) defined as the sediment density the following relationship can be determined:

\[ G=\frac{\rho_{s}}{\rho_{f}} \] \[ \rho_{s}=G \rho_{f} \] \[ \rho_{s}=2.65\cdot 1000 \] \[ \rho_{s}=2,650 (kg/m³) \]

Then, the sediment concentration by mass per unit volume (\(c_{mass}\)) is given by:

\[ c_{mass}=\rho_{s}c \]

\[ c_{mass}=2,650\cdot 0.1 \] \[ c_{mass}=265 (kg/m³) \]

Sediment concentration by ppm by weight(\(c_{ppm}\))

\[ c_{ppm}=10^6\hat{c} \] \[ c_{ppm}=10^6\cdot0.227 \] \[ c_{ppm}=0.227 ppm \]

2)Calculate the density of the water-sediment mixture (\(\rho\)).

Solution

\[ \rho = \rho_f(1-c)+\rho_sc \] \[ \rho = 1000(1-0.1)+2,650\cdot0.1 \] \[ \rho = 900+265 \] \[ \rho = 1,165 (kg/m³) \]

 

PROBLEM 2.9. A sediment mixture has the following size distribution:

 

 

1) Determine the arithmetic and geometric mean diameters;

Solution

Let`s assume \(d_{k}\) is the representative diameter given by:

\[ d_{k} = \frac{d_{lk}+d_{uk}}{2} \]  

Where:

\(d_{lk}\) is the lower bound diameter

\(d_{uk}\) is the upper bound diameter

Applying the equation above on the diameter values of the given table and considering the percentage per size class for the representative diameter (given by the difference between the percent finer range) the following table and plot is obteined.

Table2: Sediment size distribution relatively to the representative diameter

 

Figure 1: Sediment size distribution relatively to the representative diameter

 

The arithmetic mean diameter (\(d_{m}\)):

\[ d_{m}=\frac{\sum_{n=1}^{N} p_{k}d_{k}}{100} \]

 

Where:

\(d_{m}\) is the arithmetic mean diameter

\(p_{k}\) is the percentenge per size

\(N\) number of classes

\(k\) is the class index

 

Applying the equation above to the values of Table 2:

\[ d_{m}=\frac{\sum_{n=1}^{N=7} p_{k}d_{k}}{100} \]

\[ d_{m}= \frac{(5 \cdot 0.0025 + 10 \cdot 0.007 +28 \cdot 0.036 \cdot + 19.5 \cdot 0.156 +17.5 \cdot 0.625 + 10 \cdot 13 + 10 \cdot 62.5)} {100} \] \[ d_{m}= 7.701 (mm) \]

 

The geometric mean diameter (\(d_{g}\)):

\[ d_{g}= d_{1}^{p_1/100} \cdot d_{2}^{p_2/100} \cdot...d_{N}^{p_N/100} \]

\[ d_{g}= 0.0025^{5/100} \cdot 0.007^{10/100} \cdot 0.036^{28/100}\cdot 0.156^{19.5/100} \cdot 0.625^{17.5/100}\cdot 13^{10/100}cdot 62.5^{10/100} \]

\[ d_{g}= 0.223 (mm) \]

 

2) Determine the standard deviation (\(\sigma_g\)) and gradation coefficient (\(G_r\));

Solution

The standard deviation (\(\sigma\)) of the natural logarithm of the representative diameters is determined by:

\[ \sigma=\sqrt\frac{\sum_{n=1}^{N} [ln(d_k)-ln(d_g)]^2}{(N-1)} \]  

Where:

\(d_{k}\) is the representative diameter;

\(d_{g}\) is the geometric mean diameter;

\(N\) number of classes;

\(k\) class index;

Applying the equation above and the natural logarithm on the representative diameters (\(d_k\)) presented in Table 2:

 

\[ \sigma=\sqrt\frac{\sum_{n=1}^{N=7}(ln(d_{k})-ln(0.223))^2}{(7-1)} \]

 

\[ \sigma=3.763 \]  

The standard deviation of sediment size \(\sigma_g\) is determined by:

\[ \sigma_g=\Bigg(\frac{d_{84.1}}{d_{15.9}}\Bigg)^{1/2} \]  

Assuming \(d_{50}\)=\(d_{g}\), one can use the following relations to compute \(d_{15.9}\) and \(d_{84.1}\)

 

\[ ln (d_{15.9})=ln(d_{50})-\sigma \]

\[ d_{15.9}=Exp(ln(0.223)-3.763) \] \[ d_{15.9}=0.005 (mm) \]  

\[ ln (d_{84.1})=ln(d_{50})+\sigma \] \[ d_{84.1}=Exp(ln(0.223)+3.763) \] \[ d_{84.1}=9.606 \]

 

Then:

\[ \sigma_g=\Bigg(\frac{9.606}{0.005}\Bigg)^{1/2} \]

\[ \sigma_g=43.84 (mm) \]

 

The Gradation Coefficient (\(G_r\)) is determined by:

\[ G_r=\frac{1}{2}\Bigg(\frac{d_{84.1}}{d_{50}}+\frac{d_{50}}{d_{15.9}}\Bigg) \] \[ G_r=\frac{1}{2}\Bigg(\frac{9.606}{0.223}+\frac{0.223}{0.005}\Bigg) \] \[ G_r=43.84 (mm) \]

3)Calculate the porosity using the Colby method, Komura’s formula and Wu and Wang’s formula.

Solution

Before applying the Colby method we should compute the initial porosity (\(\phi_k\)) considering the representative diameter showed in Table 2. The initial porosity is given by:

\[ \phi_k=1-0.525\Bigg(\frac{d_k}{(d_k+4\delta_1)}\Bigg)^3; d_k<1mm \]  

Or

\[ \phi_k=0.3+0.175e^{-0.095(d_k-d_0)/d_0}; d_k\geq1mm \]  

Where:

\(d_{k}\) is the representative diameter;

\(\delta_{1}\) is assumed to be 0.0004 mm (water film);

\(d_0\) is assumed to be 1 mm (reference size);

\(k\) class index;

 

Applying the equations above brings to the \(\phi_k\) values presented in Table 3, as follows:

 

 

Table 3: Initial porosity

 

The overall mixture porosity by Colby method is given by (\(\phi\)):

\[ \frac{1}{(1-\phi)}=\sum_{n=1}^{N=7}\frac{p_k}{1-\phi_k} \]

 

Applying the equation above over the values presented in Table 3, brings to:

 

\[ \frac{1}{(1-\phi)}=2.398 \]

\[ \frac{1}{2.398}=(1-\phi) \]

\[ \phi=\Bigg(1-\frac{1}{2.398}\Bigg) \]

\[ \phi=0.58 \]  

Assuming, \(d_{50}\)=\(d_{g}\)=0.223:

 

The Komura empirical formula is given by

\[ \phi=0.245+\frac{0.0864}{(0.1d_{50})^{0.21}} \] \[ \phi=0.245+\frac{0.0864}{(0.1\cdot0.223)^{0.21}} \] \[ \phi=0.44 \]  

The Wu and Wang formula is given by:

\[ \phi=0.13+\frac{0.21}{(d_{50}+0.002)^{0.21}} \]

\[ \phi=0.13+\frac{0.21}{(0.223+0.002)^{0.21}} \]

\[ \phi=0.42 \]

 

PROBLEM 2.14. For sediments with size 1 mm and 20 mm, determine their angles of repose using the diagram of Simons. Assume average particle roundness.

Solution

Considering the Simons diagram and the sediments sizes of 1 mm and 20 mm and assuming an average particle roundness the repose angles are approximately 31 and 37 degrees, as shown bellow:

 

Figure 2: Simons diagram

 

PROBLEM 2.15. Consider a mixture of sand and gravel. Why and how the porosity of the mixture varies with the fraction of sand?

Solution

In multiple sized sediment deposit, the porosity is highly related to the packing or filling modes between the particles. Particle filling occurs when fine particles are finer than the voids of coarse particles. Considering a bimodal sediment mixture composed of a coarse size class (such as gravel) and a fine size class (for instance, sand) the porosity will be related with the portion of the coarse size class whose voids are being filled by fine particles. The ideal coarse packing model states that when the volume of fine particles is less than the volume of the pores of coarse particles, the fine particles are entirely used to fill in the voids between the coarse particles, and the packing mode figures as presented in Figure 3, below:

Figure 3: Coarse packing

 

When the coarse size fraction is small, the ideal fine packing model indicates that the coarse particles are completely dispersed by fine particles and lose their voids (this means B=1) and the mixture configuration looks like presented in Figure 4, as follows.

Figure 4: Fine packing

When there is no mixing or filling between the two size classes the packing configuration looks like presented in Figure 5, below.

Figure 5: No Filling

 

Wu and Li (2017) compared the mentioned models (the ideal packing model and the no filling model), the Han et al. (1981), Revised Han et al. and Wu and Li models with experimental porosity measurements conducted by Phillips (2007). The results compared how the percentage of fine particle in the mixture is related with the porosity. It is observed that increasing the fraction of fine particles until around 30%, the porosity tends to decrease. After that, increasing the percentage of fine particles brings to increasing the porosity. Figure 6 below show the described behavior, considering the different mentioned models.

Figure 6: Measured and calculated porosities for different sediment mixtures in the experiment of Phillips (2007) for a mixture of medium sand and pebble (Wu and Li, 2017)